Three consecutive even integers are such that the square of the third is 76 more than the square of the second . Find the three integers

Answer 1

Let the integers b 2n-2, 2n and 2n+2

Then (2n + 2)² = 4n² + 76

Isn't it nice that the squared terms go away and you are left with 4n = 72?

Yes, you should be able to expand the left side and then collect terms.

Then solve for n and get your 3 integers.